Appendix VL
1085
Referring to Fig. 959. Let a load of w Ibs. per foot run be
advanced from the left up to any point on the beam. It is required to
find the shearing force at the section K.
Diagram (i). The load is advanced up to K. Reaction at A has
the general form <wx (I - - j ~- / for every case. Reaction at B, which
is shear at K is wx. - -r /.
The shear at K is less than
Diagram (2). The load is short of ic.
before, because both reactions are less.
Diagram (3). When loaded beyond K. Both reactions are in-
creased ; but the shear at K is actually less than in (i) for all the load
beyond K has to be deducted in arriving at the shearing force, whereas
only part of it adds to the reaction.
Therefore the greatest shear at K occurs when the load travels
exactly up to that point.
Next take various positions for K between A and B. The ordinates
of the curve of maximum shear for these points will be, from dia-
gram (2)
. ,
wx x - -~ / =
2
r-
2/
But x is the abscissa from A
.*. Ordinates cc x*
and the curve is drawn at (2).
Pp. 446, 464, 4jo, and 885. Link Polygons.It is re-
quired to add, that is find, the resultant of a number of forces
inclined at various angles, but whose directions are co-planar,
that is, lie within the same plane. The forces may meet in one
point, and the method on p. 464 be found convenient; but if
the directions are varied in any possible co-planar manner, the
construction known as the link or funicular polygon must be
adopted.
Let the lines ABODE, Fig. 960, represent forces in direction,
sense, and magnitude whose resultant is required. Draw the
unclosed polygon o i 2 3 4 5 by lines parallel to the forces, whose
lengths are also proportional to their magnitudes. The closing
line R will naturally shew the magnitude and direction of the
resultant, and the sense will be reversed, as shewn by the arrow.
We must next shew on the space diagram the exact position of R.
Choose any pole O in the vector figure, either in or out of the